1. The problem is to find the exact value of $\sin\left(-\frac{\pi}{12}\right)$.\n\n2. We use the identity for sine of a negative angle: $$\sin(-x) = -\sin(x)$$\n\n3. So, $$\sin\left(-\frac{\pi}{12}\right) = -\sin\left(\frac{\pi}{12}\right)$$\n\n4. Next, we find $\sin\left(\frac{\pi}{12}\right)$. We can use the angle difference formula: $$\sin(a - b) = \sin a \cos b - \cos a \sin b$$\n\n5. Let $a = \frac{\pi}{4}$ and $b = \frac{\pi}{6}$, so $$\frac{\pi}{12} = \frac{\pi}{4} - \frac{\pi}{6}$$\n\n6. Calculate each term: $$\sin\frac{\pi}{4} = \frac{\sqrt{2}}{2}, \quad \cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}, \quad \cos\frac{\pi}{4} = \frac{\sqrt{2}}{2}, \quad \sin\frac{\pi}{6} = \frac{1}{2}$$\n\n7. Substitute into the formula: $$\sin\left(\frac{\pi}{12}\right) = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4}$$\n\n8. Therefore, $$\sin\left(-\frac{\pi}{12}\right) = -\frac{\sqrt{6} - \sqrt{2}}{4} = \frac{\sqrt{2} - \sqrt{6}}{4}$$\n\nFinal answer: $$\sin\left(-\frac{\pi}{12}\right) = \frac{\sqrt{2} - \sqrt{6}}{4}$$